Number of ways to go to goal vertex from start vertex
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I want to find out the number of ways to reach out to goal vertex from start vertex. The problem scenario is depicted as in the picture :
Here I want to go G vertex from S vertex. The only condition here is I can go up and right direction. DIrection diagonally/left/down is not allowed.
So keeping this condition in mind, what is the number of ways to go G vertex from S vertex?
My trying :
I have figured out the no of ways are, $1^2+2^2+3^3+ldots+(n-2)^2$. Is this correct?
sequences-and-series number-theory
edited Dec 31 '18 at 15:50
saulspatz
18k31536
asked Dec 31 '18 at 15:44
Standard EquationStandard Equation
247113
$endgroup$
add a comment |
$begingroup$
I want to find out the number of ways to reach out to goal vertex from start vertex. The problem scenario is depicted as in the picture :
Here I want to go G vertex from S vertex. The only condition here is I can go up and right direction. DIrection diagonally/left/down is not allowed.
So keeping this condition in mind, what is the number of ways to go G vertex from S vertex?
My trying :
I have figured out the no of ways are, $1^2+2^2+3^3+ldots+(n-2)^2$. Is this correct?
sequences-and-series number-theory
edited Dec 31 '18 at 15:50
saulspatz
18k31536
asked Dec 31 '18 at 15:44
Standard EquationStandard Equation
247113
$endgroup$
2
$begingroup$
can you explain how did you get the sum of squares formula?
$endgroup$
– Siong Thye Goh
Dec 31 '18 at 15:46
$begingroup$
hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
$endgroup$
– dmtri
Dec 31 '18 at 15:51
add a comment |
$begingroup$
I want to find out the number of ways to reach out to goal vertex from start vertex. The problem scenario is depicted as in the picture :
Here I want to go G vertex from S vertex. The only condition here is I can go up and right direction. DIrection diagonally/left/down is not allowed.
So keeping this condition in mind, what is the number of ways to go G vertex from S vertex?
My trying :
I have figured out the no of ways are, $1^2+2^2+3^3+ldots+(n-2)^2$. Is this correct?
sequences-and-series number-theory
edited Dec 31 '18 at 15:50
saulspatz
18k31536
asked Dec 31 '18 at 15:44
Standard EquationStandard Equation
247113
$endgroup$
I want to find out the number of ways to reach out to goal vertex from start vertex. The problem scenario is depicted as in the picture :
Here I want to go G vertex from S vertex. The only condition here is I can go up and right direction. DIrection diagonally/left/down is not allowed.
So keeping this condition in mind, what is the number of ways to go G vertex from S vertex?
My trying :
I have figured out the no of ways are, $1^2+2^2+3^3+ldots+(n-2)^2$. Is this correct?
sequences-and-series number-theory
sequences-and-series number-theory
edited Dec 31 '18 at 15:50
saulspatz
18k31536
asked Dec 31 '18 at 15:44
Standard EquationStandard Equation
247113
edited Dec 31 '18 at 15:50
saulspatz
18k31536
asked Dec 31 '18 at 15:44
Standard EquationStandard Equation
247113
edited Dec 31 '18 at 15:50
saulspatz
18k31536
edited Dec 31 '18 at 15:50
saulspatz
18k31536
edited Dec 31 '18 at 15:50
saulspatz
18k31536
18k31536
asked Dec 31 '18 at 15:44
Standard EquationStandard Equation
247113
asked Dec 31 '18 at 15:44
Standard EquationStandard Equation
247113
asked Dec 31 '18 at 15:44
Standard EquationStandard Equation
247113
247113
2
$begingroup$
can you explain how did you get the sum of squares formula?
$endgroup$
– Siong Thye Goh
Dec 31 '18 at 15:46
$begingroup$
hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
$endgroup$
– dmtri
Dec 31 '18 at 15:51
add a comment |
2
$begingroup$
can you explain how did you get the sum of squares formula?
$endgroup$
– Siong Thye Goh
Dec 31 '18 at 15:46
$begingroup$
hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
$endgroup$
– dmtri
Dec 31 '18 at 15:51
2
2
$begingroup$
can you explain how did you get the sum of squares formula?
$endgroup$
– Siong Thye Goh
Dec 31 '18 at 15:46
$begingroup$
can you explain how did you get the sum of squares formula?
$endgroup$
– Siong Thye Goh
Dec 31 '18 at 15:46
$begingroup$
hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
$endgroup$
– dmtri
Dec 31 '18 at 15:51
$begingroup$
hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
$endgroup$
– dmtri
Dec 31 '18 at 15:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.
answered Dec 31 '18 at 15:50
gt6989bgt6989b
36.1k22557
$endgroup$
$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53
1
$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29
$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.
answered Dec 31 '18 at 15:50
gt6989bgt6989b
36.1k22557
$endgroup$
$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53
1
$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29
$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56
add a comment |
$begingroup$
In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.
answered Dec 31 '18 at 15:50
gt6989bgt6989b
36.1k22557
$endgroup$
$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53
1
$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29
$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56
add a comment |
$begingroup$
In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.
answered Dec 31 '18 at 15:50
gt6989bgt6989b
36.1k22557
$endgroup$
In your case there is a direct counting approach. Your path is a sequence of $n$ right steps (R's) and $m$ up steps (U's), with $m=n=3$. So you are asking how many strings can be made from 3 R's and 3 U's, which should be $binom{6}{3}$.
answered Dec 31 '18 at 15:50
gt6989bgt6989b
36.1k22557
edited Dec 31 '18 at 16:55
answered Dec 31 '18 at 15:50
gt6989bgt6989b
36.1k22557
answered Dec 31 '18 at 15:50
gt6989bgt6989b
36.1k22557
answered Dec 31 '18 at 15:50
gt6989bgt6989b
36.1k22557
36.1k22557
$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53
1
$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29
$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56
add a comment |
$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53
1
$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29
$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56
$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53
$begingroup$
So for n = 4 , the no of ways are $ binom{8}{4} $ = 70 ?
$endgroup$
– Standard Equation
Dec 31 '18 at 15:53
1
1
$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29
$begingroup$
Isn't it $n-1$ R's and $m-1$ U's?
$endgroup$
– saulspatz
Dec 31 '18 at 16:29
$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56
$begingroup$
@saulspatz thanks fixed
$endgroup$
– gt6989b
Dec 31 '18 at 16:56
add a comment |
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$begingroup$
can you explain how did you get the sum of squares formula?
$endgroup$
– Siong Thye Goh
Dec 31 '18 at 15:46
$begingroup$
hint: how many steps you need to go up (U), right (R)? combinations between them...UURURR, UUURRR....
$endgroup$
– dmtri
Dec 31 '18 at 15:51